Don't be scared by my words, you already came across this method in 7th grade when you studied polynomials.
For example, if you needed:
Let's group: the first and third terms, as well as the second and fourth.
It is clear that the first and third are the difference of squares:
and the second and fourth have a common factor of three:
Then the original expression is equivalent to this:
Where to derive the common factor is no longer difficult:
Hence,
This is roughly what we will do when solving exponential equations: look for “commonality” among the terms and take it out of brackets, and then - come what may, I believe that we will be lucky =))
Example No. 14
The right is far from a power of seven (I checked!) And the left is not much better...
You can, of course, “chop off” the factor a from the second term from the first term, and then deal with what you got, but let’s be more prudent with you.
I don't want to deal with the fractions that inevitably form when "selecting" , so shouldn't I rather take it out?
Then I won’t have any fractions: as they say, the wolves are fed and the sheep are safe:
Calculate the expression in brackets.
Magically, magically, it turns out that (surprisingly, although what else should we expect?).
Then we reduce both sides of the equation by this factor. We get: , from.
Here's a more complicated example (quite a bit, really):
What a problem! We don't have one common ground here!
It's not entirely clear what to do now.
Let’s do what we can: first, move the “fours” to one side, and the “fives” to the other:
Now let's take out the "general" on the left and right:
So what now?
What is the benefit of such a stupid group? At first glance it is not visible at all, but let's look deeper:
Well, now we’ll make sure that on the left we only have the expression c, and on the right - everything else.
How do we do this?
Here's how: Divide both sides of the equation first by (so we get rid of the exponent on the right), and then divide both sides by (so we get rid of the numeric factor on the left).
Finally we get:
Incredible!
On the left we have an expression, and on the right we have a simple expression.
Then we immediately conclude that
Example No. 15
I will give his brief solution (without bothering myself much with explanations), try to understand all the “subtleties” of the solution yourself.
Now for the final consolidation of the material covered.
Independently solving the following 7 problems (with answers)
- Let's take the common factor out of brackets: Where:
- Let's present the first expression in the form: , divide both sides by and get that
- , then the original equation is transformed to the form: Well, now a hint - look for where you and I have already solved this equation!
- Imagine how, how, ah, well, then divide both sides by, so you get the simplest exponential equation.
- Bring it out of the brackets.
- Bring it out of the brackets.
EXPONENTARY EQUATIONS. AVERAGE LEVEL
I assume that after reading the first article, which talked about what are exponential equations and how to solve them, you have mastered the necessary minimum knowledge necessary to solve the simplest examples.
Now I will look at another method for solving exponential equations, this is...
Method for introducing a new variable (or replacement)
He solves most of the “difficult” problems on the topic of exponential equations (and not only equations).
This method is one of most often used in practice. First, I recommend that you familiarize yourself with the topic.
As you already understood from the name, the essence of this method is to introduce such a change of variable that your exponential equation will miraculously transform into one that you can easily solve.
All that remains for you after solving this very “simplified equation” is to make a “reverse replacement”: that is, return from the replaced to the replaced.
Let's illustrate what we just said with a very simple example:
Example 16. Simple replacement method
This equation can be solved using "simple replacement", as mathematicians disparagingly call it.
In fact, the replacement here is the most obvious. One has only to see that
Then the original equation will turn into this:
If you additionally imagine how, then it is absolutely clear that it is necessary to replace...
Of course, .
What then becomes the original equation? Here's what:
You can easily find its roots on your own: .
What should we do now?
It's time to return to the original variable.
What did I forget to mention?
Namely: when replacing a certain degree with a new variable (that is, when replacing a type), I will be interested in only positive roots!
You yourself can easily answer why.
Thus, you and I are not interested, but the second root is quite suitable for us:
Then where from.
Answer:
As you can see, in the previous example, a replacement was just asking for our hands. Unfortunately, this is not always the case.
However, let’s not go straight to the sad stuff, but let’s practice with one more example with a fairly simple replacement
Example 17. Simple replacement method
It is clear that most likely it will have to be replaced (this is the smallest of the degrees included in our equation).
However, before introducing a replacement, our equation needs to be “prepared” for it, namely: , .
Then you can replace, as a result I get the following expression:
Oh horror: a cubic equation with absolutely terrible formulas for solving it (well, speaking in general terms).
But let’s not despair right away, but let’s think about what we should do.
I'll suggest cheating: we know that to get a “beautiful” answer, we need to get it in the form of some power of three (why would that be, eh?).
Let's try to guess at least one root of our equation (I'll start guessing with powers of three).
First guess. Not a root. Alas and ah...
.
The left side is equal.
Right part: !
Eat! Guessed the first root. Now things will get easier!
Do you know about the “corner” division scheme? Of course you do, you use it when you divide one number by another.
But few people know that the same can be done with polynomials.
There is one wonderful theorem:
Applying to my situation, this tells me that it is divisible without remainder by.
How is division carried out? That's how:
I look to see which monomial I should multiply by to get
It is clear that on, then:
I subtract the resulting expression from, I get:
Now, what do I need to multiply by to get?
It is clear that on, then I will get:
and again subtract the resulting expression from the remaining one:
Well, the last step is to multiply by and subtract from the remaining expression:
Hurray, division is over! What have we accumulated in private?
By itself: .
Then we got the following expansion of the original polynomial:
Let's solve the second equation:
It has roots:
Then the original equation:
has three roots:
We will, of course, discard the last root, since it is less than zero.
And the first two after reverse replacement will give us two roots:
Answer: ..
I didn’t mean to scare you with this example!
Rather, on the contrary, my goal was to show that although we had a fairly simple replacement, it nevertheless led to a rather complex equation, the solution of which required some special skills from us.
Well, no one is immune from this. But the replacement in this case was quite obvious.
Example No. 18 (with a less obvious replacement)
It is not at all clear what we should do: the problem is that in our equation there are two different bases and one foundation cannot be obtained from another by raising it to any (reasonable, naturally) degree.
However, what do we see?
Both bases differ only in sign, and their product is the difference of squares equal to one:
Definition:
Thus, the numbers that are the bases in our example are conjugate.
In this case, the smart step would be multiply both sides of the equation by the conjugate number.
For example, on, then left side the equation will become equal, and the right one.
If we make a substitution, then our original equation will become like this:
its roots, then, and remembering that, we get that.
Answer: , .
As a rule, the replacement method is sufficient to solve most “school” exponential equations.
The following tasks of an increased level of complexity are taken from the Unified State Exam variants.
Three tasks of increased complexity from the Unified State Exam variants
You are already literate enough to solve these examples on your own. I will only give the required replacement.
- Solve the equation:
- Find the roots of the equation:
- Solve the equation: . Find all the roots of this equation that belong to the segment:
And now some brief explanations and answers:
Example No. 19
Here it is enough for us to note that...
Then the original equation will be equivalent to this:
This equation can be solved by replacing
Do the further calculations yourself.
In the end, your task will be reduced to solving simple trigonometric problems (depending on sine or cosine). We will look at solutions to similar examples in other sections.
Example No. 20
Here you can even do without replacement...
It is enough to move the subtrahend to the right and represent both bases through powers of two: , and then immediately move on to the quadratic equation.
Example No. 21
This is also solved in a fairly standard way: let’s imagine how.
Then, replacing, we get a quadratic equation: then,
You already know what a logarithm is, right? No? Then read the topic urgently!
The first root obviously does not belong to the segment, but the second one is unclear!
But we will find out very soon!
Since, then (this is a property of the logarithm!)
Subtract from both sides, then we get:
The left side can be represented as:
multiply both sides by:
can be multiplied by, then
Then compare:
since then:
Then the second root belongs to the required interval
Answer:
As you see, selection of roots of exponential equations requires a fairly deep knowledge of the properties of logarithms, so I advise you to be as careful as possible when solving exponential equations.
As you understand, in mathematics everything is interconnected!
As my math teacher said: “mathematics, like history, cannot be read overnight.”
As a rule, all The difficulty in solving problems of an increased level of complexity is precisely the selection of the roots of the equation.
Another example for practice...
Example 22
It is clear that the equation itself is solved quite simply.
By making a substitution, we reduce our original equation to the following:
First let's look at first root.
Let's compare and: since, then. (property logarithmic function, at).
Then it is clear that the first root does not belong to our interval.
Now the second root: . It is clear that (since the function at is increasing).
It remains to compare and...
since, then, at the same time.
This way I can “drive a peg” between the and.
This peg is a number.
The first expression is less and the second is greater.
Then the second expression is greater than the first and the root belongs to the interval.
Answer: .
Finally, let's look at another example of an equation where the substitution is quite unusual.
Example No. 23 (Equation with non-standard replacement!)
Let's start right away with what can be done, and what - in principle, can be done, but it is better not to do it.
You can imagine everything through the powers of three, two and six.
Where it leads?
It won’t lead to anything: a jumble of degrees, some of which will be quite difficult to get rid of.
What then is needed?
Let's notice that a
And what will this give us?
And the fact that we can reduce the solution of this example to the solution of a fairly simple exponential equation!
First, let's rewrite our equation as:
Now let's divide both sides of the resulting equation by:
Eureka! Now we can replace, we get:
Well, now it’s your turn to solve demonstration problems, and I will give only brief comments to them so that you don’t go astray! Good luck!
Example No. 24
The most difficult!
It’s so hard to see a replacement here! But nevertheless, this example can be completely solved using highlighting a complete square.
To solve it, it is enough to note that:
Then here's your replacement:
(Note that here in our replacement we cannot discard negative root!!! Why do you think?)
Now to solve the example you only have to solve two equations:
Both of them can be solved by a “standard replacement” (but the second one in one example!)
Example No. 25
2. Notice that and make a replacement.
Example No. 26
3. Decompose the number into coprime factors and simplify the resulting expression.
Example No. 27
4. Divide the numerator and denominator of the fraction by (or, if you prefer) and make the substitution or.
Example No. 28
5. Notice that the numbers and are conjugate.
SOLVING EXPONENTARY EQUATIONS USING THE LOGARIFHM METHOD. ADVANCED LEVEL
In addition, let's look at another way - solving exponential equations using the logarithm method.
I can’t say that solving exponential equations using this method is very popular, but in some cases only it can lead us to the right decision our equation.
It is especially often used to solve the so-called “ mixed equations": that is, those where functions of different types occur.
Example No. 29
V general case can only be solved by taking the logarithm of both sides (for example, to the base), which will transform the original equation into the following:
Let's look at the following example:
It is clear that according to the ODZ of the logarithmic function, we are only interested.
However, this follows not only from the ODZ of the logarithm, but for one more reason.
I think it won’t be difficult for you to guess which one it is.
Let's take the logarithm of both sides of our equation to the base:
As you can see, taking the logarithm of our original equation quickly led us to the correct (and beautiful!) answer.
Let's practice with one more example.
Example No. 30
There’s nothing wrong here either: let’s take the logarithm of both sides of the equation to the base, then we get:
Let's make a replacement:
However, we missed something! Did you notice where I made a mistake? After all, then:
which does not satisfy the requirement (think where it came from!)
Answer:
Try to write down the solution to the exponential equations below:
Now compare your decision with this:
Example No. 31
Let's logarithm both sides to the base, taking into account that:
(the second root is not suitable for us due to replacement)
Example No. 32
Let's take logarithms to the base:
Let us transform the resulting expression to the following form:
EXPONENTARY EQUATIONS. BRIEF DESCRIPTION AND BASIC FORMULAS
Exponential equation
Equation of the form:
called the simplest exponential equation.
Properties of degrees
Approaches to solution
- Reduction to the same basis
- Reduction to the same exponent
- Variable replacement
- Simplifying the expression and applying one of the above.
Go to the youtube channel of our website to stay up to date with all the new video lessons.
First, let's remember the basic formulas of powers and their properties.
Product of a number a occurs on itself n times, we can write this expression as a a … a=a n
1. a 0 = 1 (a ≠ 0)
3. a n a m = a n + m
4. (a n) m = a nm
5. a n b n = (ab) n
7. a n / a m = a n - m
Power or exponential equations– these are equations in which the variables are in powers (or exponents), and the base is a number.
Examples of exponential equations:
IN in this example the number 6 is the base, it is always at the bottom, and the variable x degree or indicator.
Let us give more examples of exponential equations.
2 x *5=10
16 x - 4 x - 6=0
Now let's look at how exponential equations are solved?
Let's take a simple equation:
2 x = 2 3
This example can be solved even in your head. It can be seen that x=3. After all, in order for the left and right sides to be equal, you need to put the number 3 instead of x.
Now let’s see how to formalize this decision:
2 x = 2 3
x = 3
In order to solve such an equation, we removed identical grounds(that is, twos) and wrote down what was left, these are degrees. We got the answer we were looking for.
Now let's summarize our decision.
Algorithm for solving the exponential equation:
1. Need to check the same whether the equation has bases on the right and left. If the reasons are not the same, we are looking for options to solve this example.
2. After the bases become the same, equate degrees and solve the resulting new equation.
Now let's look at a few examples:
Let's start with something simple.
The bases on the left and right sides are equal to the number 2, which means we can discard the base and equate their powers.
x+2=4 The simplest equation is obtained.
x=4 – 2
x=2
Answer: x=2
In the following example you can see that the bases are different: 3 and 9.
3 3x - 9 x+8 = 0
First, move the nine to the right side, we get:
Now you need to make the same bases. We know that 9=3 2. Let's use the power formula (a n) m = a nm.
3 3x = (3 2) x+8
We get 9 x+8 =(3 2) x+8 =3 2x+16
3 3x = 3 2x+16 now you can see that in the left and right side the bases are the same and equal to three, which means we can discard them and equate the degrees.
3x=2x+16 we get the simplest equation
3x - 2x=16
x=16
Answer: x=16.
Let's look at the following example:
2 2x+4 - 10 4 x = 2 4
First of all, we look at the bases, bases two and four. And we need them to be the same. We transform the four using the formula (a n) m = a nm.
4 x = (2 2) x = 2 2x
And we also use one formula a n a m = a n + m:
2 2x+4 = 2 2x 2 4
Add to the equation:
2 2x 2 4 - 10 2 2x = 24
We gave an example for the same reasons. But other numbers 10 and 24 bother us. What to do with them? If you look closely you can see that on the left side we have 2 2x repeated, here is the answer - we can put 2 2x out of brackets:
2 2x (2 4 - 10) = 24
Let's calculate the expression in brackets:
2 4 — 10 = 16 — 10 = 6
We divide the entire equation by 6:
Let's imagine 4=2 2:
2 2x = 2 2 bases are the same, we discard them and equate the degrees.
2x = 2 is the simplest equation. Divide it by 2 and we get
x = 1
Answer: x = 1.
Let's solve the equation:
9 x – 12*3 x +27= 0
Let's transform:
9 x = (3 2) x = 3 2x
We get the equation:
3 2x - 12 3 x +27 = 0
Our bases are the same, equal to three. In this example, you can see that the first three has a degree twice (2x) than the second (just x). In this case, you can solve replacement method. We replace the number with the smallest degree:
Then 3 2x = (3 x) 2 = t 2
We replace all x powers in the equation with t:
t 2 - 12t+27 = 0
We get a quadratic equation. Solving through the discriminant, we get:
D=144-108=36
t 1 = 9
t2 = 3
Returning to the variable x.
Take t 1:
t 1 = 9 = 3 x
That is,
3 x = 9
3 x = 3 2
x 1 = 2
One root was found. We are looking for the second one from t 2:
t 2 = 3 = 3 x
3 x = 3 1
x 2 = 1
Answer: x 1 = 2; x 2 = 1.
On the website you can ask any questions you may have in the HELP DECIDE section, we will definitely answer you.
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At the stage of preparation for the final test, high school students need to improve their knowledge on the topic “ Exponential equations" The experience of past years indicates that such tasks cause certain difficulties for schoolchildren. Therefore, high school students, regardless of their level of preparation, need to thoroughly master the theory, remember the formulas and understand the principle of solving such equations. Having learned to cope with this type of problem, graduates can count on high scores when passing the Unified State Exam in mathematics.
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When reviewing the materials they have covered, many students are faced with the problem of finding the formulas needed to solve equations. A school textbook is not always at hand, and selecting the necessary information on a topic on the Internet takes a long time.
The Shkolkovo educational portal invites students to use our knowledge base. We are implementing a completely new method of preparing for the final test. By studying on our website, you will be able to identify gaps in knowledge and pay attention to those tasks that cause the most difficulty.
Shkolkovo teachers collected, systematized and presented all the material necessary for successfully passing the Unified State Exam in the simplest and most accessible form.
Basic definitions and formulas are presented in the “Theoretical background” section.
To better understand the material, we recommend that you practice completing the assignments. Carefully review the examples of exponential equations with solutions presented on this page to understand the calculation algorithm. After that, proceed to perform tasks in the “Directories” section. You can start with the easiest tasks or go straight to solving complex exponential equations with several unknowns or . The database of exercises on our website is constantly supplemented and updated.
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Solving exponential equations. Examples.
Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)
What's happened exponential equation? This is an equation in which the unknowns (x's) and expressions with them are in indicators some degrees. And only there! It is important.
There you are examples of exponential equations:
3 x 2 x = 8 x+3
Note! In the bases of degrees (below) - only numbers. IN indicators degrees (above) - a wide variety of expressions with an X. If, suddenly, an X appears in the equation somewhere other than an indicator, for example:
this will be an equation mixed type. Such equations do not have clear rules for solving them. We will not consider them for now. Here we will deal with solving exponential equations in its purest form.
In fact, even pure exponential equations are not always solved clearly. But there are certain types of exponential equations that can and should be solved. These are the types we will consider.
Solving simple exponential equations.
First, let's solve something very basic. For example:
Even without any theories, by simple selection it is clear that x = 2. Nothing more, right!? No other value of X works. Now let's look at the solution to this tricky exponential equation:
What have we done? We, in fact, simply threw out the same bases (triples). Completely thrown out. And, the good news is, we hit the nail on the head!
Indeed, if in an exponential equation there are left and right the same numbers in any powers, these numbers can be removed and the exponents can be equalized. Mathematics allows. It remains to solve a much simpler equation. Great, right?)
However, let us remember firmly: You can remove bases only when the base numbers on the left and right are in splendid isolation! Without any neighbors and coefficients. Let's say in the equations:
2 x +2 x+1 = 2 3, or
twos cannot be removed!
Well, we have mastered the most important thing. How to move from evil exponential expressions to simpler equations.
"Those are the times!" - you say. “Who would give such a primitive lesson on tests and exams!?”
I have to agree. Nobody will. But now you know where to aim when solving tricky examples. It must be brought to the form where the same base number is on the left and right. Then everything will be easier. Actually, this is a classic of mathematics. We take the original example and transform it to the desired one us mind. According to the rules of mathematics, of course.
Let's look at examples that require some additional effort to reduce them to the simplest. Let's call them simple exponential equations.
Solving simple exponential equations. Examples.
When solving exponential equations, the main rules are actions with degrees. Without knowledge of these actions nothing will work.
To actions with degrees, one must add personal observation and ingenuity. Do we need the same base numbers? So we look for them in the example in explicit or encrypted form.
Let's see how this is done in practice?
Let us be given an example:
2 2x - 8 x+1 = 0
The first keen glance is at grounds. They... They are different! Two and eight. But it’s too early to become discouraged. It's time to remember that
Two and eight are relatives in degree.) It is quite possible to write:
8 x+1 = (2 3) x+1
If we recall the formula from operations with degrees:
(a n) m = a nm ,
this works out great:
8 x+1 = (2 3) x+1 = 2 3(x+1)
The original example began to look like this:
2 2x - 2 3(x+1) = 0
We transfer 2 3 (x+1) to the right (no one has canceled the elementary operations of mathematics!), we get:
2 2x = 2 3(x+1)
That's practically all. Removing the bases:
We solve this monster and get
This is the correct answer.
In this example, knowing the powers of two helped us out. We identified in eight there is an encrypted two. This technique (encryption of common grounds under different numbers) is a very popular technique in exponential equations! Yes, and in logarithms too. You must be able to recognize powers of other numbers in numbers. This is extremely important for solving exponential equations.
The fact is that raising any number to any power is not a problem. Multiply, even on paper, and that’s it. For example, anyone can raise 3 to the fifth power. 243 will work out if you know the multiplication table.) But in exponential equations, much more often it is not necessary to raise to a power, but vice versa... Find out what number to what degree is hidden behind the number 243, or, say, 343... No calculator will help you here.
You need to know the powers of some numbers by sight, right... Let's practice?
Determine what powers and what numbers the numbers are:
2; 8; 16; 27; 32; 64; 81; 100; 125; 128; 216; 243; 256; 343; 512; 625; 729, 1024.
Answers (in a mess, of course!):
5 4 ; 2 10 ; 7 3 ; 3 5 ; 2 7 ; 10 2 ; 2 6 ; 3 3 ; 2 3 ; 2 1 ; 3 6 ; 2 9 ; 2 8 ; 6 3 ; 5 3 ; 3 4 ; 2 5 ; 4 4 ; 4 2 ; 2 3 ; 9 3 ; 4 5 ; 8 2 ; 4 3 ; 8 3 .
If you look closely you can see strange fact. There are significantly more answers than tasks! Well, it happens... For example, 2 6, 4 3, 8 2 - that's all 64.
Let us assume that you have taken note of the information about familiarity with numbers.) Let me also remind you that to solve exponential equations we use all stock of mathematical knowledge. Including those from junior and middle classes. You didn’t go straight to high school, right?)
For example, when solving exponential equations, putting the common factor out of brackets often helps (hello to 7th grade!). Let's look at an example:
3 2x+4 -11 9 x = 210
And again, the first glance is at the foundations! The bases of the degrees are different... Three and nine. But we want them to be the same. Well, in this case the desire is completely fulfilled!) Because:
9 x = (3 2) x = 3 2x
Using the same rules for dealing with degrees:
3 2x+4 = 3 2x ·3 4
That’s great, you can write it down:
3 2x 3 4 - 11 3 2x = 210
We gave an example for the same reasons. So, what is next!? You can't throw out threes... Dead end?
Not at all. Remember the most universal and powerful decision rule everyone math assignments:
If you don’t know what you need, do what you can!
Look, everything will work out).
What's in this exponential equation Can do? Yes, on the left side it just begs to be taken out of brackets! The overall multiplier of 3 2x clearly hints at this. Let's try, and then we'll see:
3 2x (3 4 - 11) = 210
3 4 - 11 = 81 - 11 = 70
The example keeps getting better and better!
We remember that to eliminate grounds we need a pure degree, without any coefficients. The number 70 bothers us. So we divide both sides of the equation by 70, we get:
Oops! Everything got better!
This is the final answer.
It happens, however, that taxiing on the same basis is achieved, but their elimination is not possible. This happens in other types of exponential equations. Let's master this type.
Replacing a variable in solving exponential equations. Examples.
Let's solve the equation:
4 x - 3 2 x +2 = 0
First - as usual. Let's move on to one base. To a deuce.
4 x = (2 2) x = 2 2x
We get the equation:
2 2x - 3 2 x +2 = 0
And this is where we hang out. The previous techniques will not work, no matter how you look at it. We'll have to pull out another powerful and universal method from our arsenal. It's called variable replacement.
The essence of the method is surprisingly simple. Instead of one complex icon (in our case - 2 x) we write another, simpler one (for example - t). Such a seemingly meaningless replacement leads to amazing results!) Everything just becomes clear and understandable!
So let
Then 2 2x = 2 x2 = (2 x) 2 = t 2
In our equation we replace all powers with x's by t:
Well, is it dawning on you?) Quadratic equations Have you forgotten yet? Solving through the discriminant, we get:
The main thing here is not to stop, as happens... This is not the answer yet, we need x, not t. Let's return to the X's, i.e. we make a reverse replacement. First for t 1:
That is,
One root was found. We are looking for the second one from t 2:
Hm... 2 x on the left, 1 on the right... Problem? Not at all! It is enough to remember (from operations with powers, yes...) that a unit is any number to the zero power. Any. Whatever is needed, we will install it. We need a two. Means:
That's it now. We got 2 roots:
This is the answer.
At solving exponential equations at the end sometimes you end up with some kind of awkward expression. Type:
From seven to two through simple degree does not work. They are not relatives... How can we be? Someone may be confused... But the person who read on this site the topic “What is a logarithm?” , just smile sparingly and write down with a steady hand absolutely correct answer:
There cannot be such an answer in tasks “B” on the Unified State Examination. There a specific number is required. But in tasks “C” it’s easy.
This lesson provides examples of solving the most common exponential equations. Let's highlight the main points.
1. First of all, we look at grounds degrees. We are wondering if it is possible to make them identical. Let's try to do this by actively using actions with degrees. Don't forget that numbers without x's can also be converted to powers!
2. We try to bring the exponential equation to the form when on the left and on the right there are the same numbers in any powers. We use actions with degrees And factorization. What can be counted in numbers, we count.
3. If the second tip doesn’t work, try using variable replacement. The result may be an equation that can be easily solved. Most often - square. Or fractional, which also reduces to square.
4. To successfully solve exponential equations, you need to know the powers of some numbers by sight.
As usual, at the end of the lesson you are invited to decide a little.) On your own. From simple to complex.
Solve exponential equations:
More difficult:
2 x+3 - 2 x+2 - 2 x = 48
9 x - 8 3 x = 9
2 x - 2 0.5x+1 - 8 = 0
Find the product of roots:
2 3's + 2 x = 9
Happened?
Well, then a very complex example (though it can be solved in the mind...):
7 0.13x + 13 0.7x+1 + 2 0.5x+1 = -3
What's more interesting? Then here's a bad example for you. Quite tempting for increased difficulty. Let me hint that in this example, what saves you is ingenuity and the most universal rule for solving all mathematical problems.)
2 5x-1 3 3x-1 5 2x-1 = 720 x
A simpler example, for relaxation):
9 2 x - 4 3 x = 0
And for dessert. Find the sum of the roots of the equation:
x 3 x - 9x + 7 3 x - 63 = 0
Yes Yes! This is a mixed type equation! Which we did not consider in this lesson. Why consider them, they need to be solved!) This lesson is quite enough to solve the equation. Well, you need ingenuity... And may seventh grade help you (this is a hint!).
Answers (in disarray, separated by semicolons):
1; 2; 3; 4; there are no solutions; 2; -2; -5; 4; 0.
Is everything successful? Great.
There is a problem? No problem! Special Section 555 solves all these exponential equations with detailed explanations. What, why, and why. And, of course, there is additional valuable information on working with all sorts of exponential equations. Not just these ones.)
One last fun question to consider. In this lesson we worked with exponential equations. Why didn’t I say a word about ODZ here? In equations, this is a very important thing, by the way...
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
Back forward
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Lesson type
: lesson on generalization and complex application of knowledge, skills and abilities on the topic “Exponential equations and methods for solving them.”Lesson objectives.
Equipment:
computer and multimedia projector.Used in class information Technology : methodological support to the lesson - presentation in Microsoft Power Point.
During the classes
Every skill comes with hard work
I. Setting a lesson goal(Slide number 2 )
In this lesson, we will summarize and generalize the topic “Exponential equations, their solutions.” Let's get acquainted with typical Unified State Exam assignments different years on this topic.
Problems on solving exponential equations can be found in any part of the Unified State Examination tasks. In the part “ IN " Usually they offer to solve the simplest exponential equations. In the part “ WITH " You can find more complex exponential equations, the solution of which is usually one of the stages of completing the task.
For example ( Slide number 3 ).
- Unified State Examination - 2007
Q 4 – Find the greatest value of the expression x y, Where ( X; at) – solution of the system:
Q 1 – Solve the equations:
A) X 6 3X – 36 6 3X = 0;
b) 4 X +1 + 8 4X= 3.
Q 4 – Find the meaning of the expression x + y, Where ( X; at) – solution of the system:
- Unified State Examination - 2010
II. Updating basic knowledge. Repetition
(Slides No. 4 – 6 presentations for the lesson)Shown on screen background summary of theoretical material on this topic.
The following issues are discussed:
- What equations are called indicative?
- Name the main ways to solve them. Give examples of their types ( Slide number 4 )
- What theorem is used when solving simple exponential equations of the form: and f(x) = a g(x) ?
- What other methods for solving exponential equations exist? ( Slide number 5 )
(Independently solve the proposed equations for each method and perform a self-test using the slide)
- Factorization method (based on properties of powers with identical grounds, technique: the degree with the lowest indicator is taken out of brackets).
- Method of division (multiplication) by an exponential expression other than zero when solving homogeneous exponential equations .
- Advice:
-
Solving equations using the last two methods with subsequent comments
(Slide number 6 ).
. 4 X+ 1 – 2 4 X– 2 = 124, 4 X– 2 (4 3 - 2) = 124, 4 X– 2 62 = 124,4 X– 2 = 2, 4 X– 2 = 4 0,5 , X– 2 = 0,5, x = 2,5 .
2 2 2х – 3 2 X 5X - 5 5 2X= 0¦: 5 2 X 0,2 (2/5) 2х – 3 (2/5) X - 5 = 0,
t = (2/5) x, t > 0, 2t 2 - 3t- 5 = 0,t= -1(?...), t = 5/2; 5/2 = (2/5) x, X= ?...
III. Solving Unified State Exam 2010 tasks
Students independently solve the tasks proposed at the beginning of the lesson on slide No. 3, using instructions for the solution, check their progress in solving and answers to them using a presentation ( Slide number 7). During the work, options and solutions are discussed, attention is drawn to possible mistakes when deciding.
: a) 7 X– 2 = 49, b) (1/6) 12 – 7 x = 36. Answer: A) X= 4, b) X = 2. : 4 X 2 + 3X – 2 - 0,5 2x2 + 2X– 1 = 0. (Can be replaced by 0.5 = 4 – 0.5)Solution. ,
X 2 + 3X – 2 = -X 2 - 4X + 0,5 …
Answer: X= -5/2, X = 1/2.
: 5 5 tg y+ 4 = 5 -tg y, at cos y< 0.Directions to the solution
. 5 5 tg y+ 4 = 5 -tg y¦ 5 tg y 0,5 5 2g y+ 4 5 tg y – 1 = 0. Let X= 5 tg y , …
5 tg y = -1 (?...), 5 tg y = 1/5.
Since tg y= -1 and cos y< 0, then at II coordinate quarter
Answer: at= 3/4 + 2k, k N.
IV. Teamwork at the board
A high level of training task is being considered - Slide number 8. With the help of this slide, a dialogue between the teacher and students occurs, facilitating the development of a solution.
– At what parameter A equation 2 2 X – 3 2 X + A 2 – 4A= 0 has two roots?Let t= 2 X, Where t > 0 . We get t 2 – 3t + (A 2 – 4A) = 0 .
1). Since the equation has two roots, then D > 0;
2). Because t 1,2 > 0, then t 1 t 2 > 0, that is A 2 – 4A> 0 (?...).
Answer: A(– 0.5; 0) or (4; 4.5).
V. Test work
(Slide number 9 )Students perform test work on pieces of paper, exercising self-monitoring and self-evaluation of the work performed using a presentation, becoming established in the topic. They independently determine for themselves a program for regulating and correcting knowledge based on mistakes made in workbooks. Sheets with completed independent work are handed over to the teacher for checking.
Underlined numbers – basic level, with an asterisk – increased complexity.
Solution and answers.
3. 2 X– 1 (5 2 4 - 4) = 19, 2 X– 1 76 = 19, 2 X– 1 = 1/4, 2 X– 1 = 2 – 2 , X– 1 = -2,
x = -1.
4 *.3 9 x = 2 3 X 5X+ 5 25 X | : 25 X ,
3 (9/25) x = 2 (3/5) X+ 5,
3 (9/27) X = 2 (3/5) X + 5 = 0,
3 (3/5) 2X – 2 (3/5) X - 5 = 0,…, (3/5) X = -1 (doesn't fit),
(3/5) X = 5, x = -1.
VI. Homework assignment
(Slide number 10 )- Repeat § 11, 12.
- From the materials of the Unified State Exam 2008 - 2010, select tasks on the topic and solve them.
- Home test work :
